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<meta property="og:description" content="等差数列划分如果一个数列至少有三个元素，并且任意两个相邻元素之差相同，则称该数列为等差数列。 例如，以下数列为等差数列: 1, 3, 5, 7, 97, 7, 7, 73, -1, -5, -9 以下数列不是等差数列。 1, 1, 2, 5, 7 1数组 A 包含 N 个数，且索引从0开始。数组 A 的一个子数组划分为数组 (P, Q)，P 与 Q 是整数且满足 0&lt;&#x3D;P&lt;Q&lt;N">
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fa-fw"></i></a></div></div></nav><div id="post-info"><h1 class="post-title">Leetcode 413.等差数列划分</h1><div id="post-meta"><div class="meta-firstline"><span class="post-meta-date"><i class="far fa-calendar-alt fa-fw post-meta-icon"></i><span class="post-meta-label">发表于</span><time class="post-meta-date-created" datetime="2019-12-25T03:21:09.000Z" title="发表于 2019-12-25 11:21:09">2019-12-25</time><span class="post-meta-separator">|</span><i class="fas fa-history fa-fw post-meta-icon"></i><span class="post-meta-label">更新于</span><time class="post-meta-date-updated" datetime="2021-03-30T12:41:35.763Z" title="更新于 2021-03-30 20:41:35">2021-03-30</time></span><span class="post-meta-categories"><span class="post-meta-separator">|</span><i class="fas fa-inbox fa-fw post-meta-icon"></i><a class="post-meta-categories" href="/categories/%E7%AE%97%E6%B3%95%E9%A2%98/">算法题</a></span></div><div class="meta-secondline"><span class="post-meta-separator">|</span><span class="post-meta-wordcount"><i class="far fa-file-word fa-fw post-meta-icon"></i><span class="post-meta-label">字数总计:</span><span class="word-count">1.2k</span><span class="post-meta-separator">|</span><i class="far fa-clock fa-fw post-meta-icon"></i><span class="post-meta-label">阅读时长:</span><span>5分钟</span></span><span class="post-meta-separator">|</span><span class="post-meta-pv-cv" id="" data-flag-title="Leetcode 413.等差数列划分"><i class="far fa-eye fa-fw post-meta-icon"></i><span class="post-meta-label">阅读量:</span><span id="busuanzi_value_page_pv"></span></span></div></div></div></header><main class="layout" id="content-inner"><div id="post"><article class="post-content" id="article-container"><h1 id="等差数列划分"><a href="#等差数列划分" class="headerlink" title="等差数列划分"></a>等差数列划分</h1><p>如果一个数列至少有三个元素，并且任意两个相邻元素之差相同，则称该数列为等差数列。</p>
<p>例如，以下数列为等差数列:</p>
<p>1, 3, 5, 7, 9<br>7, 7, 7, 7<br>3, -1, -5, -9</p>
<p>以下数列不是等差数列。</p>
<p>1, 1, 2, 5, 7</p>
<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">数组 A 包含 N 个数，且索引从<span class="number">0</span>开始。数组 A 的一个子数组划分为数组 (P, Q)，P 与 Q 是整数且满足 <span class="number">0</span>&lt;=P&lt;Q&lt;N 。</span><br></pre></td></tr></table></figure>
<p>如果满足以下条件，则称子数组(P, Q)为等差数组：<br><figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">元素 A[P], A[p + <span class="number">1</span>], ..., A[Q - <span class="number">1</span>], A[Q] 是等差的。并且 P + <span class="number">1</span> &lt; Q 。</span><br></pre></td></tr></table></figure><br>函数要返回数组 A 中所有为等差数组的子数组个数。</p>
<p>示例:</p>
<p>A = [1, 2, 3, 4]</p>
<p>返回: 3, A 中有三个子等差数组: [1, 2, 3], [2, 3, 4] 以及自身 [1, 2, 3, 4]。</p>
<h1 id="思路-代码"><a href="#思路-代码" class="headerlink" title="思路 + 代码"></a>思路 + 代码</h1><p>首先对于等差序列 B, 其元素数量为n，则其包含的连续自等差序列的总数为 1+2+…+n-2,<br>因此，该题转化为寻找序列中，<strong>最长的连续子等差序列</strong>，然后根据其数量判断。</p>
<figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span>:</span></span><br><span class="line">    <span class="function"><span class="keyword">def</span> <span class="title">numberOfArithmeticSlices</span>(<span class="params">self, A: List[<span class="built_in">int</span>]</span>) -&gt; int:</span></span><br><span class="line">        l = <span class="built_in">len</span>(A)</span><br><span class="line">        <span class="keyword">if</span> l &lt; <span class="number">3</span>: <span class="keyword">return</span> <span class="number">0</span></span><br><span class="line">        res, count = <span class="number">0</span>, <span class="number">0</span></span><br><span class="line">        <span class="keyword">for</span> i <span class="keyword">in</span> <span class="built_in">range</span>(<span class="number">2</span>,l):</span><br><span class="line">            <span class="keyword">if</span> A[i]-A[i-<span class="number">1</span>] == A[i-<span class="number">1</span>]-A[i-<span class="number">2</span>]:</span><br><span class="line">                count+=<span class="number">1</span></span><br><span class="line">            <span class="keyword">else</span>:</span><br><span class="line">                <span class="keyword">if</span> count!=<span class="number">0</span>:</span><br><span class="line">                    res += <span class="built_in">sum</span>(<span class="built_in">range</span>(count+<span class="number">1</span>))</span><br><span class="line">                    count=<span class="number">0</span></span><br><span class="line">        <span class="keyword">if</span> count != <span class="number">0</span>:</span><br><span class="line">            res += <span class="built_in">sum</span>(<span class="built_in">range</span>(count + <span class="number">1</span>))</span><br><span class="line">        <span class="keyword">return</span> res</span><br></pre></td></tr></table></figure>
<p>来源：力扣（LeetCode）<br>链接：<a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/arithmetic-slices">https://leetcode-cn.com/problems/arithmetic-slices</a><br>著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。</p>
<h1 id="题目"><a href="#题目" class="headerlink" title="题目"></a>题目</h1><p>给定一个正整数 n，将其拆分为至少两个正整数的和，并使这些整数的乘积最大化。 返回你可以获得的最大乘积。</p>
<p>示例 1:<br><figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">输入: <span class="number">2</span></span><br><span class="line">输出: <span class="number">1</span></span><br><span class="line">解释: <span class="number">2</span> = <span class="number">1</span> + <span class="number">1</span>, <span class="number">1</span> × <span class="number">1</span> = <span class="number">1</span>。</span><br></pre></td></tr></table></figure><br>示例 2:<br><figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">输入: <span class="number">10</span></span><br><span class="line">输出: <span class="number">36</span></span><br><span class="line">解释: <span class="number">10</span> = <span class="number">3</span> + <span class="number">3</span> + <span class="number">4</span>, <span class="number">3</span> × <span class="number">3</span> × <span class="number">4</span> = <span class="number">36</span>。</span><br></pre></td></tr></table></figure></p>
<p>说明: 你可以假设 n 不小于 2 且不大于 58。</p>
<h1 id="思路-代码-1"><a href="#思路-代码-1" class="headerlink" title="思路 + 代码"></a>思路 + 代码</h1><p>动态规划，整数4的最大乘积为: dp[3] = max(max(dp[2], dp[1]<em>2), 1 </em> 2)</p>
<figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span>:</span></span><br><span class="line">    <span class="function"><span class="keyword">def</span> <span class="title">integerBreak</span>(<span class="params">self, n: <span class="built_in">int</span></span>) -&gt; int:</span></span><br><span class="line">        <span class="keyword">if</span> n &lt; <span class="number">2</span>: <span class="keyword">return</span> <span class="number">0</span></span><br><span class="line">        dp = [<span class="number">1</span>] * (n+<span class="number">1</span>)</span><br><span class="line">        <span class="keyword">for</span> i <span class="keyword">in</span> <span class="built_in">range</span>(<span class="number">2</span>, n + <span class="number">1</span>):</span><br><span class="line">            <span class="keyword">for</span> j <span class="keyword">in</span> <span class="built_in">range</span>(<span class="number">1</span>, i):</span><br><span class="line">                dp[i] = <span class="built_in">max</span>(dp[j] * (i - j), dp[i])</span><br><span class="line">                <span class="comment"># 很关键的一步，因为之前的dp[i]最大值可能比dp[i]小，例如2或者3</span></span><br><span class="line">                dp[i] = <span class="built_in">max</span>(j*(i-j), dp[i])</span><br><span class="line">        <span class="keyword">return</span> dp[-<span class="number">1</span>]</span><br></pre></td></tr></table></figure>
<p>来源：力扣（LeetCode）<br>链接：<a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/integer-break">https://leetcode-cn.com/problems/integer-break</a><br>著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。</p>
<h1 id="完全平方数"><a href="#完全平方数" class="headerlink" title="完全平方数"></a>完全平方数</h1><p>动态规划， 与前一题类似</p>
<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">import</span> java.lang.Math;</span><br><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> </span>&#123;</span><br><span class="line">    <span class="function"><span class="keyword">public</span> <span class="keyword">int</span> <span class="title">numSquares</span><span class="params">(<span class="keyword">int</span> n)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">if</span>(n&lt;=<span class="number">3</span>) <span class="keyword">return</span> n;</span><br><span class="line">        <span class="keyword">int</span>[] dp = <span class="keyword">new</span> <span class="keyword">int</span>[n+<span class="number">1</span>];</span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">0</span>;i&lt;n+<span class="number">1</span>; i++)&#123;</span><br><span class="line">            dp[i] = Integer.MAX_VALUE;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>; i&lt;=<span class="number">3</span>;i++)&#123;</span><br><span class="line">            dp[i] = i;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">4</span>; i&lt;n+<span class="number">1</span>; i++)&#123;</span><br><span class="line">            <span class="keyword">int</span> max_n = (<span class="keyword">int</span>)Math.sqrt(i);</span><br><span class="line">            <span class="keyword">if</span>(max_n*max_n==i)&#123;</span><br><span class="line">                dp[i]=<span class="number">1</span>;</span><br><span class="line">            &#125; <span class="keyword">else</span>&#123;</span><br><span class="line">                <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">1</span>; j&lt;=max_n; j++)&#123;</span><br><span class="line">                    dp[i] = Math.min(dp[i], dp[i-j*j]+dp[j*j]);</span><br><span class="line">                &#125;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> dp[n];</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span>(<span class="params"><span class="built_in">object</span></span>):</span></span><br><span class="line">    <span class="function"><span class="keyword">def</span> <span class="title">numSquares</span>(<span class="params">self, n</span>):</span></span><br><span class="line">        <span class="string">&quot;&quot;&quot;</span></span><br><span class="line"><span class="string">        :type n: int</span></span><br><span class="line"><span class="string">        :rtype: int</span></span><br><span class="line"><span class="string">        &quot;&quot;&quot;</span></span><br><span class="line">        <span class="keyword">if</span> n &lt;= <span class="number">3</span>:</span><br><span class="line">            <span class="keyword">return</span> n</span><br><span class="line">        dp = [sys.maxsize] * (n + <span class="number">1</span>)</span><br><span class="line">        <span class="keyword">for</span> i <span class="keyword">in</span> <span class="built_in">range</span>(<span class="number">1</span>, <span class="number">4</span>):</span><br><span class="line">            dp[i] = i</span><br><span class="line">        <span class="keyword">for</span> i <span class="keyword">in</span> <span class="built_in">range</span>(<span class="number">4</span>, n + <span class="number">1</span>):</span><br><span class="line">            max_n = <span class="built_in">int</span>(math.sqrt(i))</span><br><span class="line">            <span class="keyword">if</span> max_n * max_n == i:</span><br><span class="line">                dp[i] = <span class="number">1</span></span><br><span class="line">            <span class="keyword">else</span>:</span><br><span class="line">                <span class="keyword">for</span> j <span class="keyword">in</span> <span class="built_in">range</span>(<span class="number">1</span>, max_n + <span class="number">1</span>):</span><br><span class="line">                    dp[i] = <span class="built_in">min</span>(dp[i], dp[i - j * j] + dp[j*j])</span><br><span class="line">        <span class="keyword">return</span> dp[-<span class="number">1</span>]</span><br></pre></td></tr></table></figure>
<figure class="highlight javascript"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">/**</span></span><br><span class="line"><span class="comment"> * <span class="doctag">@param <span class="type">&#123;number&#125;</span> <span class="variable">n</span></span></span></span><br><span class="line"><span class="comment"> * <span class="doctag">@return <span class="type">&#123;number&#125;</span></span></span></span><br><span class="line"><span class="comment"> */</span></span><br><span class="line"><span class="keyword">var</span> numSquares = <span class="function"><span class="keyword">function</span>(<span class="params">n</span>) </span>&#123;</span><br><span class="line">    <span class="keyword">if</span>(n&lt;=<span class="number">3</span>) <span class="keyword">return</span> n;</span><br><span class="line">    <span class="keyword">let</span> dp = <span class="keyword">new</span> <span class="built_in">Array</span>(n+<span class="number">1</span>).fill(<span class="built_in">Number</span>.MAX_VALUE);</span><br><span class="line">    <span class="keyword">for</span>(<span class="keyword">let</span> i=<span class="number">1</span>; i&lt;=<span class="number">3</span>; i++)&#123;</span><br><span class="line">        dp[i] = i;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">for</span>(<span class="keyword">let</span> i=<span class="number">4</span>; i&lt;n+<span class="number">1</span>; i++)&#123;</span><br><span class="line">        max_n = <span class="built_in">Math</span>.trunc(<span class="built_in">Math</span>.sqrt(i));</span><br><span class="line">        <span class="keyword">if</span>(max_n*max_n==i) dp[i]=<span class="number">1</span>;</span><br><span class="line">        <span class="keyword">else</span>&#123;</span><br><span class="line">            <span class="keyword">for</span>(<span class="keyword">let</span> j=<span class="number">1</span>; j&lt;=max_n; j++)&#123;</span><br><span class="line">                dp[i] = <span class="built_in">Math</span>.min(dp[i], dp[i-j*j]+dp[j*j]);</span><br><span class="line">            &#125;</span><br><span class="line">        &#125; </span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> dp[n];</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>
<h1 id="解码方法"><a href="#解码方法" class="headerlink" title="解码方法"></a>解码方法</h1><p>一条包含字母 A-Z 的消息通过以下方式进行了编码：</p>
<p>‘A’ -&gt; 1<br>‘B’ -&gt; 2<br>…<br>‘Z’ -&gt; 26<br>给定一个只包含数字的非空字符串，请计算解码方法的总数。</p>
<p>示例 1:</p>
<p>输入: “12”<br>输出: 2<br>解释: 它可以解码为 “AB”（1 2）或者 “L”（12）。<br>示例 2:</p>
<p>输入: “226”<br>输出: 3<br>解释: 它可以解码为 “BZ” (2 26), “VF” (22 6), 或者 “BBF” (2 2 6) 。</p>
<p><strong>思路：</strong>动态规划，需注意0的处理，1010的编码方式共有1种，而909编码方式为0种，202编码方式为1种。</p>
<figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span>(<span class="params"><span class="built_in">object</span></span>):</span></span><br><span class="line">    <span class="function"><span class="keyword">def</span> <span class="title">numDecodings</span>(<span class="params">self, s</span>):</span></span><br><span class="line">        <span class="string">&quot;&quot;&quot;</span></span><br><span class="line"><span class="string">        :type s: str</span></span><br><span class="line"><span class="string">        :rtype: int</span></span><br><span class="line"><span class="string">        &quot;&quot;&quot;</span></span><br><span class="line">        l = <span class="built_in">len</span>(s)</span><br><span class="line">        <span class="keyword">if</span> l==<span class="number">0</span> <span class="keyword">or</span> s[<span class="number">0</span>]==<span class="string">&#x27;0&#x27;</span>: <span class="keyword">return</span> <span class="number">0</span></span><br><span class="line">        dp = [<span class="number">0</span>]*(l+<span class="number">1</span>)</span><br><span class="line">        dp[<span class="number">0</span>], dp[<span class="number">1</span>] = <span class="number">1</span>, <span class="number">1</span></span><br><span class="line">        <span class="keyword">for</span> i <span class="keyword">in</span> <span class="built_in">range</span>(<span class="number">2</span>, l+<span class="number">1</span>):</span><br><span class="line">            <span class="keyword">if</span> s[i-<span class="number">1</span>]!=<span class="string">&#x27;0&#x27;</span>:</span><br><span class="line">                dp[i] += dp[i-<span class="number">1</span>]</span><br><span class="line">            <span class="keyword">if</span> <span class="number">9</span>&lt;<span class="built_in">int</span>(s[i-<span class="number">2</span>:i])&lt;=<span class="number">26</span>:</span><br><span class="line">                dp[i]+=dp[i-<span class="number">2</span>]</span><br><span class="line">        <span class="keyword">return</span> dp[-<span class="number">1</span>]      </span><br></pre></td></tr></table></figure>
<p>来源：力扣（LeetCode）<br>链接：<a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/decode-ways">https://leetcode-cn.com/problems/decode-ways</a><br>著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。</p>
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href="/LeetCode-1143-%E6%9C%80%E9%95%BF%E5%85%AC%E5%85%B1%E5%AD%90%E5%BA%8F%E5%88%97/" title="LeetCode 1143.最长公共子序列"><img class="cover" src="https://cdn.jsdelivr.net/gh/SUNYunZeng/sources/img/astronaut.png" alt="cover"><div class="content is-center"><div class="date"><i class="far fa-calendar-alt fa-fw"></i> 2019-12-27</div><div class="title">LeetCode 1143.最长公共子序列</div></div></a></div></div></div><hr/><div id="post-comment"><div class="comment-head"><div class="comment-headline"><i class="fas fa-comments fa-fw"></i><span> 评论</span></div></div><div class="comment-wrap"><div><div id="waline-wrap"></div></div></div></div></div><div class="aside-content" id="aside-content"><div class="card-widget card-info"><div class="is-center"><div class="avatar-img"><img src="/img/avatar.jpg" onerror="this.onerror=null;this.src='/img/friend_404.gif'" alt="avatar"/></div><div class="author-info__name">孙云增</div><div class="author-info__description">极简生活，极致内涵</div></div><div class="card-info-data"><div 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